Given the function $$f(x,y)=x^3+3xy+y^5$$, solve for $$(df)/(dy)$$.

This question asks to solve for the partial derivative of the function $f$ with respect to $y$.

In order to calculate the partial derivative, we view $x$ as a fixed number and calculate the normal derivative with respect to $y$.

To find the derivative, we must apply the sum rule of derivatives, which states that the derivative of a sum is equal to the sum of the derivatives. Therefore, the derivative can be expressed as:

$\frac{df}{dy}=\frac{d\left(x^3\right)}{dy}+\frac{d\left(3xy\right)}{dy}+\frac{d\left(y^5\right)}{dy}$

Since we are treating x as a fixed number, the first term becomes 0 since the derivative of a constant number is equal to 0. After solving the rest of the terms, we are left with this final expression:

$\frac{df}{dy}=0 + 3x + 5y^4$

Which simplifies to:

$\frac{df}{dy}=3x + 5y^4$